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JIPMER Jipmer Medical Solved Paper-2000

  • question_answer
    When light of wavelength 300 nm falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, however light of 600 nm wavelength is sufficient for creating photoemission. What is the ratio of the work functions of the two emitters:

    A)  1 : 4                                      

    B)         4 : 1                      

    C)         2 : 1                                      

    D)         1 : 2

    Correct Answer: C

    Solution :

    From the formula of work function \[{{W}_{0}}=\frac{hc}{\lambda }\] here \[{{\lambda }_{0}}\] is threshold wavelength \[\frac{{{W}_{01}}}{{{W}_{02}}}=\frac{{{\lambda }_{02}}}{{{\lambda }_{01}}}=\frac{600}{300}=\frac{2}{1}\]


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